Find an equation of the line through the point (2, 1) that cuts off the least area
from the first quadrant.
The line must satisfy f(2) = 1, so the slope and y-intercept will vary. We want the minimum area.
$\displaystyle y=mx+b$ with $\displaystyle 1=2m+b$,
$\displaystyle b=1-2m$ and $\displaystyle y=mx+1-2m$
Therefore...
$\displaystyle y=m(x-2) + 1$
We want to minimize the area of the 1st quadrant or the area of y-intercept * x-intercept
The x-intercept is when y=0
$\displaystyle 0=m(x-2)+1$
$\displaystyle x-int=\frac{-1}{m}+2$
The y-intercept is just b
b = 1-2m
Area must be equal to: $\displaystyle 0.5(\frac{-1}{m}+2)(1-2m) = 0.5(\frac{-1}{m} + 2 + 2 - 4m)$
Find the derivative and set equal to zero...
$\displaystyle 0 = \frac{0.5}{m^2} - 2$
$\displaystyle 2m^2 = 0.5$
$\displaystyle m^2 = 0.25$
$\displaystyle m = -0.5, .5$
Slope must equal -0.5 since a positive slope will give an infinite area in the 1st quadrant...
AREA: $\displaystyle 0.5(\frac{-1}{m}+2)(1-2m) = 0.5(\frac{-1}{-0.5} + 2 + 2 - 4(-0.5))$
$\displaystyle 0.5(\frac{-1}{-0.5} + 2 + 2 - 4(-0.5)) = 4$
Good up to this point. The y-intercept is 1 - 2m as you showed above.
So the area will be
$\displaystyle A = \frac{1}{2} \left ( -\frac{1}{m} + 2 \right )(1 - 2m)$
By the way, it changes nothing in the answer, but you should at least deal with what happens in the case of a vertical line or horizontal line, just to be complete.
-Dan